Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 59 / 2 / e / Solution

The meaningful twistor-geometric interpretation of “colinear in $PN$” is that the entire projective line through two distinct projective null twistors lies in projective null twistor space. Mere membership of two points in a projective line would impose no condition, because every pair of distinct points in ${\mathbb{CP}}^3$ spans one.

Indeed, individual nullness givesThis vanishes for every $a,b\in \mathbb{C}$ precisely when $h(X,Y)=0$. Thus $W=\mathrm{span}(X,Y)$ is a maximal totally isotropic two-plane, a real point of Minkowski conformal compactification under the Lorentzian reality structure on twistor space. The two corresponding alpha-planes meet at this unique $W$. Hence their compact intersection is a real spacetime point, and their real null geodesics meet there.

There are two qualifications. The point can be at conformal infinity, so a finite affine meeting is not automatic. If $X,Y$ represent the same projective point instead, their alpha-planes coincide rather than having a unique intersection. If “colinear” is read only as spanning a line in ${\mathbb{CP}}^3$, individual nullness alone does not force a real intersection; the additional orthogonality condition above is indispensable.

For a concrete counterexample to the weaker reading, take $X=(0,0,1,0)$ and $Y=(1,0,0,1)$ in $({\omega }^0,{\omega }^1,{\pi }_{0^{\prime }},{\pi }_{1^{\prime }})$ coordinates. Each is a null twistor, but $h(X,Y)=1$. Their two incidence equations give $x=-i\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, which is not Hermitian. Thus their alpha-planes meet at a complex point that is not a real Minkowski point.